Question: Does there exist a group $H$ such that $|H/Z(H)| = 6$?
If there were to exist such a group then $H/Z(H) \cong C_6$ or $S_3$. Note that if $H/Z(H) \cong C_6$ then $H$ would be abelian, implying that $H = Z(H)$ and hence $|H/Z(H)| = 1$; contradiction. So we must have $H/Z(H)\cong S_3$ if there is to exist such a group. If we take $H$ to be $S_3$, then $Z(H) = \{e\}$ as if we have the permutation $(a_1 \space a_2 \dots a_k)$ then $(a_1 \space a_2)(a_1 \space a_2 \cdots a_k) = (a_2 \space a_3 \cdots a_k)$ but $(a_1 \space a_2 \cdots a_k)(a_1 \space a_2) = (a_1 \space a_3 \cdots a_k)$. So $|S_3 / Z(S_3)| = \frac{|S_3|}{|Z(S_3)|} = \frac{3!}{1} = 6$ and so we have found such a group.
Is this valid?
Let me summarise the comments here, so that the question has an answer. If the quotient group $H(Z(H)$ has six elements, then it must be isomorphic to either $C_6$ or $S_3$. This is well known, and a short proof can be found here: There are only two types of groups of order $6.$
Since $H/Z(H)$ cannot be cyclic, as you have shown correctly (since $H/Z(H)$ cyclic implies that $H$ is abelian), we are left with the case of finding groups $H$ such that ${\rm Inn (H)}\cong H/Z(H)\cong S_3$.
It is clear that $H=S_3$ and $H=D_6$ satisfy $H/Z(H)\cong S_3$, because $Z(S_3)=1$ and $Z(D_6)\cong C_2$. Actually, $D_6\cong C_2\times S_3$, and every direct product $H=A\times S_3$ with $A$ abelian satisfies $H/Z(H)\cong S_3$.
A new example is the dicyclic group $H=C_3\rtimes C_4$ of order $12$, with $H/Z(H)\cong S_3$.